of 17 /17
Solving Inequalities, Compound Inequalities and Absolute Value Inequalities Sec 1.5 &1.6 pg. 33 - 43

Solving Inequalities, Compound Inequalities and Absolute Value Inequalities Sec 1.5 &1.6 pg. 33 - 43

Embed Size (px)

Text of Solving Inequalities, Compound Inequalities and Absolute Value Inequalities Sec 1.5 &1.6 pg. 33 - 43

  • Solving Inequalities, Compound Inequalities and Absolute Value InequalitiesSec 1.5 &1.6

    pg. 33 - 43

    Hallford 2007Glencoe 2003

    ObjectivesThe learner will be able to (TLWBAT):solve inequalitiessolve real-world problems with inequalitiessolve compound inequalitiessolve absolute value inequalities

    Hallford 2007Glencoe 2003

    PropertiesTrichotomy PropertyFor any two real numbers a and b, a < ba = ba > b

    Hallford 2007Glencoe 2003

    Properties of InequalitiesFor any real number a, b, and cAddition Prop.If a>b, then a + c > b + cIf ab, then a c > b cIf a 2x + 2-x -x-7 > x + 2-2 -2-9 > xNow we need to graph our answer on a number line.

    Hallford 2007Glencoe 2003

    Graphing on a Number LineLets look at our previous answer, -9 > x-90What does this answer mean?Are my xs here?Or here?-9 is GREATER than x, so my xs that make senseare -10, -10.1, -11, etc. Anything smaller than -9For < or > you use an opening dot or point - For < or > you use a closed dot or point -

    Hallford 2007Glencoe 2003

    Properties of InequalitiesMultiplication Prop.For any real numbers a, b, and cIf c is positiveIf a>b, then ac>bcIf ab/cIf a -5}This is read as the set of all x such that x is greater than or equal to -5

    Hallford 2007Glencoe 2003

    Work this problem3(a +4) 2(3a +4) 4a - 13a + 12 6a 8 4a 1 -3a + 4 4a 1 4 7a - 15 7a5/7 aNow graph andexpress in setbuilder notation.0{a | a 5/7}You can also use interval notation. Interval notation uses ( & ) for < or > and [ & ] for & . We also use - (negative infinity) & + (positive infinity)Interval notation for this problem would be [5/7, +)

    Hallford 2007Glencoe 2003

    Compound InequalitiesA compound inequality consists of two inequalities joined by the word and or the word or.You must solve both inequalities and then graph.The final graph of the and inequality is the intersection of both individual solution graphs.The final graph of the or inequality in the union of both individual solution graphs

    Hallford 2007Glencoe 2003

    Lets solve an and compound inequality7 < 2x - 1 < 15Method 1 Divide into two problems7 < 2x 18 < 2x4 < xMethod 2 Work together2x 1 < 152x < 16x < 87 < 2x 1 < 158 < 2x < 164 < x < 8Whateveryou do toone sidedo to the other!Now Graph4 < xx < 8000444888{ x | 4 x < 8}Means 7 < 2x -1 and 2x 1< 154 < x < 8

    Hallford 2007Glencoe 2003

    Lets work about the same problem as an or inequality7 < 2x -1 or 2x 1 < 15We know the answer is 4 < x and x < 8, but thistime the answer graph is different. It is the UNIONof the two graphs.4 < xx < 80004488The solution set is all real numbers .4 < x or x < 8

    Hallford 2007Glencoe 2003

    Work this problem2x + 7 < -1 or 3x + 7 > 102x + 7 < -13x + 7 > 102x < -8x < -43x > 3x > 10-40or10-41

    Hallford 2007Glencoe 2003

    Absolute Value Inequalities|a| < b, where b > 0work as an and problem-b < a < b|a| > b, where b > 0work as an or problema > b or a < -b

    Hallford 2007Glencoe 2003

    Work these problems|x 1| < 3-3 < x -1 < 3-2 < x < 4|x -1 | > 3x -1 > 3 or x -1 < -3x > 4 or x < -200

    Hallford 2007Glencoe 2003

    Absolute Value Inequalities|a| < b, where b < 0if b is less than zero, it is negative, so there is no solution|a| > b, where b < 0if b is less than zero, it is negative, so every real number is a solution or all reals.

    Hallford 2007Glencoe 2003

    Work these problems|2x 7| < -5There is no solutionfor the above sincethe absolute value cannot be less than zero.|3x 1| + 9 > 2|3x 1| > -7Any value of x willmake this statementtrue, since the absolute value is alwaysgreater than a negativenumberAll Real numbers